Finding the area between two functions.

Having problems with the following question.

For the equations y=sqrt{9-x} and y=x+3

a)Find the value of x when they intersect

b)Find the area between the curves from x=-1 to x=2.

-I found a to be x=0, but i am getting b wrong, weather its my arithmetic or my steps are wrong, i don't know.

-When inputting the two functions into an integral equation, should i take them away? multiply them? or......... i have been taking them away.

If anyone could help me with this, it would be greatly appreciated, thanks.

Re: Finding the area between two functions.

Quote:

Originally Posted by

**johnsy123** -When inputting the two functions into an integral equation, should i take them away? multiply them? or......... i have been taking them away.

$\displaystyle A=\int_{-1}^2|\sqrt{9-x}-(x+3)|\;dx=$

$\displaystyle \int_{-1}^0(\sqrt{9-x}-x-3)\;dx+\int_{0}^2(x+3-\sqrt{9-x})\;dx=\ldots$

Re: Finding the area between two functions.

http://i22.photobucket.com/albums/b3...n/integral.jpg

The area you are trying to evaluate is the combined pink and blue regions.

For the pink region, $\displaystyle \displaystyle y = \sqrt{9-x}$ is the greater function, so the area is $\displaystyle \displaystyle \int_{-1}^0{\sqrt{9-x}\,dx} - \int_{-1}^0{x + 3\,dx}$.

For the blue region, $\displaystyle \displaystyle y = x + 3$ is the greater function, so the area is $\displaystyle \displaystyle \int_0^2{x + 3\,dx} - \int_0^2{\sqrt{9-x}\,dx}$.

Your required area is the sum of the resulting two answers.